\[h(\mathbf{x})=\mathbf{x}-
	\begin{pmatrix}
		\phantom{-} 1 \\-21
	\end{pmatrix}
	,\quad
	g(\mathbf{x})=
	\begin{pmatrix}
		98 & -1 \\1 & \phantom{-} 98
	\end{pmatrix}
	\mathbf{x},\quad
	h^{-1}(\mathbf{x})=\mathbf{x}+
	\begin{pmatrix}
		\phantom{-}1 \\-21
	\end{pmatrix}
	.\solnmarksplus{2}{1 for formulas for \(h\) and \(h^{-1}\),\\ 1 for formula for \(g\)\\ (can be implicit in working)}\]

Hence
\begin{align*}
	k(\mathbf{x}) & =(h^{-1}\circ g \circ h)(\mathbf{x})                       \\ & =h^{-1}(g(h(\mathbf{x})))\\
	%line3
	              & =h^{-1}\left( g \left(\mathbf{x}-
	\begin{pmatrix}
		\phantom{-} 1 \\-21
	\end{pmatrix}
	\right)\right)                                                             \\
	%line4
	              & =h^{-1}\left(
	\begin{pmatrix}
		98 & -1 \\1 & \phantom{-} 98
	\end{pmatrix}
	\left(\mathbf{x}-
	\begin{pmatrix}
		\phantom{-} 1 \\-21
	\end{pmatrix}
	\right)\right)\solnmarksplus{1}{substituting formulas in correctly}        \\
	%line5
	              & =h^{-1}\left(
	\begin{pmatrix}
		98 & -1 \\1 & \phantom{-} 98
	\end{pmatrix}
	\mathbf{x}-
	\begin{pmatrix}
		98 & -1 \\1 & \phantom{-} 98
	\end{pmatrix}
	\begin{pmatrix}
		\phantom{-} 1 \\-21
	\end{pmatrix}
	\right)                                                                    \\
	%line6
	              & =h^{-1}\left(
	\begin{pmatrix}
		98 & -1 \\1 & \phantom{-} 98
	\end{pmatrix}
	\mathbf{x}-
	\begin{pmatrix}
		3 \\1
	\end{pmatrix}
	\right)\solnmarksplus{1}{product of matrix and vector}                     \\
	%line7
\end{align*}
